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In the correctly worked addition problem above, M, N, R, S, T and V
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06 May 2015, 02:27
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In the correctly worked addition problem above, M, N, P, R, S, T and V are distinct digits. Is R > 3? (1) M, N and P are positive even integers. (2) S = 2 Attachment:
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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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06 May 2015, 09:03
Bunuel wrote: In the correctly worked addition problem above, M, N, R, S, T and V are distinct digits. Is R > 3? (1) M, N and P are positive even integers. (2) S = 2 Attachment: Alphametic.jpg Kudos for a correct solution.i think answer is B I suppose P is also distinct digit 1. M, N and P are positive even integers. [2,4,6,8] if R=3 then M 3 N 3 P 3 + _ _ 9 now take any value of M,N,P from 2,4,6,8 we will get a 3 digit(distinct) sum if R = 4 then also for any even value of M,N,P from 2,4,6,8 we will get a 3 digit(distinct) sum. Not sufficient (2) S = 2 if R = 3 take any value of M,N,P from 1,2,3,4,5,6,7,8,9 such that there sum gives a 2 digit no with S = 2 this will also get satisfied if R=4 thus not sufficient (1)+(2) S=2 and M,N,P from 2,4,6,8 only R>=7 will satisfy the condition that S(2), T, V all distinct. i.e. take M,N and P as 4,6,8 4 7 6 7 +8 7 2 0 1
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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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06 May 2015, 09:50
I think it would be C. Both statements together gives R>3 1 only does not give any information. Consider M, N and P to be 2, 4, 6 respectively. It doesn't tell anything about the value of R.
2 only is also insufficent. If R=3 and M=6 N=7 P=8 It gives S=2 Also, if M=6 N=4 P=2, then for S to be 2 R has to be greater than 3.
1+2) No combination of M, N and P as positive even integers gives the sum >=20. So, in order for S=2 the extra value should come from the sum of R+R+R i.e. 3R and for that R has to be greater than 3.



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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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06 May 2015, 14:44
Pretty sure it's D.
With statement 1, adding together any combination of three digits between 2,4,6, and 8 gives you a two digit number with the second digit being one of the already used digits from 2,4,6, and 8. For instance, 2+4+6=12. 4+6+8=18, etc. This means that the ones column, or the Rs, need to add up to be at least 10 so you can carry a 1 and avoid the redundancy between number "T" and numbers "M","N", and "P." In order to achieve that, the "R"s have to be more than 3, since 3+3+3 is 9. Therefore Statement 1 is sufficient.
With statement 2, we know that "M","N" and "P" have to add up to at least 18 (since the most we could carry over from the "R"s in the addition process would be a 2). This means that "M","N" and "P" have to either be 765, 876, or 987. 7+6+5=18, so R would have to be 7 or 9, to get 21 or 27, but both of these involve repeating numbers so they won't work. 8+7+6=21, so R can be any of the remaining numbers, but 3 is the only one that works without any redundancies. 9+8+7=24, so R can be any of the remaining numbers, but again, 1 is the only one that works without any redundancies. Neither of those possibilities are greater than 3, so statement 2 is also sufficient.



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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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06 May 2015, 17:02
[quote="Bunuel"] M, N, R, S, T and V are distinct digits. R > 3? (1) M, N and P are positive even integers M, N, P could be 2, 4, 6, or 4, 6, 8 or 2, 6, 8 If the digits of M, N, P is 2, 4, 6, then S and T would be 1 and 2. This is not possible because M is 2 and T is 2 and that goes against the "distinct integers" In fact, 4,6,8 and 2,6,8 all result in a repeated digit for T. R then must be relatively larger. in this case at least 4 in order to change the value of T to a digit that is distinct. Sufficient (2) S = 2 96+46+76=218 R>3? yes 90+80+70=240 R>3? no insufficient Answer: A



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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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06 May 2015, 20:55
fhp wrote: Pretty sure it's D.
With statement 1, adding together any combination of three digits between 2,4,6, and 8 gives you a two digit number with the second digit being one of the already used digits from 2,4,6, and 8. For instance, 2+4+6=12. 4+6+8=18, etc. This means that the ones column, or the Rs, need to add up to be at least 10 so you can carry a 1 and avoid the redundancy between number "T" and numbers "M","N", and "P." In order to achieve that, the "R"s have to be more than 3, since 3+3+3 is 9. Therefore Statement 1 is sufficient.
With statement 2, we know that "M","N" and "P" have to add up to at least 18 (since the most we could carry over from the "R"s in the addition process would be a 2). This means that "M","N" and "P" have to either be 765, 876, or 987. 7+6+5=18, so R would have to be 7 or 9, to get 21 or 27, but both of these involve repeating numbers so they won't work. 8+7+6=21, so R can be any of the remaining numbers, but 3 is the only one that works without any redundancies. 9+8+7=24, so R can be any of the remaining numbers, but again, 1 is the only one that works without any redundancies. Neither of those possibilities are greater than 3, so statement 2 is also sufficient. A quick tip  The two statements cannot give you contradictory answers. They are a part of the same puzzle. If the question is "what is x?" Stmnt 1 cannot give you x = 4 if stmnt 2 gives you x = 5. If you get contradictory answers, it means you have messed up somewhere. Here, apparently, stmnt 1 gives you that R is greater than 3 while stmnt 2 gives you that R cannot be greater than 3  these contradict. So you must have messed up in one of the statements. Go back to see which one. Interestingly, four solutions above give four different answers!
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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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07 May 2015, 00:18
Correct answer is D. Here is my explanation: (1) M, N, P are even integers and cannot be 0 => M, N, P can be any of the set A {2, 4, 6, 8}. In any case M + N + P > 10. _ Assumption: R < or = 3 => 3R < 10 > there is no 1 carried to the tenth > M + N + P is even (sum of 3 even integers is an even integer). So far we know that M, N, P, T are 4 different even integers and cannot be 0 (sum of any 3 even integers in set A above is > 10). _ Note that sum M + N + P always contains 10 (2+8 or 4+6) > the unit digit of sum M + N + P will be M/N/P but not T > contrast to the question stem. => Assumption is wrong => R > 3 : Sufficient. (2) S = 2 => M + N + P + (3R / 10) >= 20. Because Max(M + N + P) = 4 + 6 + 8=18 => 3R > 20 => R > 3 : Sufficient. Answer: D Hope you guys can understand my answer



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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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07 May 2015, 23:53
Its E because even f we combine both the statements we will be able to get both R>3 and R< 3



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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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08 May 2015, 12:12
Hi Bunuel, Can you confirm that the question is written correctly: Is "P" supposed to be included in the list of distinct digits or can it be a DUPLICATE of one of them? GMAT assassins aren't born, they're made, Rich
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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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10 May 2015, 20:23
EMPOWERgmatRichC wrote: Hi Bunuel,
Can you confirm that the question is written correctly:
Is "P" supposed to be included in the list of distinct digits or can it be a DUPLICATE of one of them?
GMAT assassins aren't born, they're made, Rich P is also supposed to be in the list of distinct digits  all digits are distinct. I missed it in the original question.
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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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11 May 2015, 03:29



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In the correctly worked addition problem above, M, N, R, S, T and V
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11 May 2015, 04:47
Bunuel wrote: In the correctly worked addition problem above, M, N, P, R, S, T and V are distinct digits. Is R > 3? (1) M, N and P are positive even integers. (2) S = 2 Attachment: The attachment Alphametic.jpg is no longer available Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: This is certainly harder than the PS question but our process will remain the same. First, let’s see what information we are given in the question – the units digits of all three numbers are the same. The twodigit numbers add up to give a three digit number. The hundreds digit, S, is either 1 or 2. Three twodigit numbers cannot add up to give a number 300 or more since 99 + 99 + 99 = 297. We have no information on what the value of R can be. All we know is that R cannot be 0 because 0+0+0 = 0 but V needs to be different from R. Let’s look at the statements now. Statement 1: M, N and P are positive even integers. At first, it may seem that this has nothing to do with the value of R but we must analyze what is given to be sure. M, N and P must take distinct values out of 2, 4, 6 and 8 and add up to give the units digit of T (again, distinct) Every time you add three even numbers, you will get an even number. Let’s see which combinations we can get: 2 + 4 + 6 = 12 2 + 4 + 8 = 14 2 + 6 + 8 = 16 4 + 6 + 8 = 18 Note that in all four cases, the units digit is one of the numbers but T must be distinct. This means that there must have been a carryover from the previous addition. So when we added the three Rs, we must have got a carryover. Had R been 3 or less, we would not have got a carryover since 1+1+1 = 3, 2+2+2 = 6 and 3+3+3 = 9. So R must be greater than 3. One such case would be Attachment:
Alphametic1.jpg [ 3.21 KiB  Viewed 3014 times ]
This statement alone is sufficient. Statement 2: S = 2 The result of addition gives us a number which is more than 200. In statement 1 we saw a case in which S is 2 and R is greater than 3. Now all we have to do is find a case in which S is 2 and R is less than 3. One of these cases is Attachment:
Alphametic2.jpg [ 3.18 KiB  Viewed 3015 times ]
So this statement alone is not sufficient. Answer (A)
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Re: In the correctly worked addition problem above, M, N, R, S, T and V
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10 Jun 2018, 21:27
Let's start by analysing the first statement It is given that M,N and R are positive even integers. We have 4 options for this: 2,4,6,8 The sum of any three of them will give a repeat digit. For eg: 2+4+6= 12 (t=2) This is not possible as M and T must be distinct 4+6+8= 18 (t=8) This is not possible as M and T must be distinct For T to be different from M,N and R, it must be odd. This will happen if we have a carryover from R+R+R=3R 3R>10 for the carryover hence R>3. Sufficient.
(2) S=2 This is not sufficient because we get S=2 with values of R greater than 3 and less than 3.




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